If the random variable X follows the Poisson distribution such that $P[X = k] = P[X = k+1]$, then the mean value of X is:

$k+1$

The correct answer is Option (4) → $k+1$

Given: $X \sim \text{Poisson}(\lambda)$ and $P[X = k] = P[X = k+1]$

Poisson pmf: $P[X = k] = \frac{e^{-\lambda} \lambda^k}{k!}$

Given equality:

$\frac{e^{-\lambda} \lambda^k}{k!} = \frac{e^{-\lambda} \lambda^{k+1}}{(k+1)!}$

Cancel $e^{-\lambda}$:

$\frac{\lambda^k}{k!} = \frac{\lambda^{k+1}}{(k+1)!}$

Simplify:

$\frac{\lambda^k}{k!} = \frac{\lambda \lambda^k}{(k+1)k!} \Rightarrow 1 = \frac{\lambda}{k+1} \Rightarrow \lambda = k+1$

Mean of Poisson = $\lambda = k+1$